Exercise 10.3.5 (Unique subgroup of cyclic group)

Question

Refresh your memory by proving this fact about
subgroups of cyclic groups: the cyclic group of 
order $n$ has exactly one subgroup
of order $k$ for each divisor $k$ of $n$.

Hassaan Naeem · Accepted Answer

extbf{Solution:}
Let $C_n$ be a cyclic group of order $n$ and let $C_k$ be a subgroup of order $k$.
If $k|n$ then we have that $n = kq$.
We also have that $C_n = \langle g angle$, and hence $C_k = \langle g^q angle$.
Suppose we also have another subgroup $C_m 
eq 1$ such that $|C_m| = k$.
We then let $r$ be the least number, $1,2,3,...$, such that $g^r \in C_m$, in other words it is the
smallest non-identity element (0) of the group. 
Since $|C_m| = k$, we then must have that $g^{rk} = 1$.
This is because the identity is $g^0$, hence the $k$-th element is $g^{k-1}$, then
due to the cyclicity of the group $g^{rk} = 1$. Similarly, $g^n = 1$. Hence $n|rk$ and we have that $rk = nm = kqm \implies r=qm$,
hence $m|r$. Hence, $g^r = g^{qm}$, since $g^q \in C_k$ then $g^{qm} \in C_k$.
Hence $C_k = C_m$.

Richard Ganaye · Answer

\begin{proof} 
Let $G = \langle g angle$ be a cyclic group of order $n$, and $k>0$ a divisor of $n$, so that $n = kq$ for some integer $q$. Since $g$ has order $n$, $g^q$ has order $k = n/q$: indeed, for all integers $s$,

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$$(g^q)^s =e \iff n \mid qs \iff \frac{n}{q} \mid s \iff k \mid s.$$
Therefore $H_0 = \langle g^q angle$ is a subgroup of $G$ of order $k$.

\bigskip
Now, let $H$ be any subgroup of order $k$. Since $g^n =e \in H$, then $A = \{u \in \N^* \mid g^u \in H\} 
e \varnothing$, hence there is a smallest positive integer $r$ such that $g^r \in H$:
$$r = \min(A) = \min\{u \in \N^* \mid g^u \in H\}.$$
Then $h = g^r \in H$. Moreover, if $x$ is any element of $H$, then $x = g^s$ for some integer $s$. The Euclidean division gives $s =qr + r_0$, where $0 \leq r_0 < r$. Then $g^{r_0} = g^s (g^r)^{-q} \in H$ (because $g^s = x \in H$, and $h = g^r \in H$). If $r_0 
e 0$, then $r_0 \in A$, and $r_0 < \min(A)$: this is a contradiction, thus $r_0 = 0$. Therefore $x = g^s = (g^r)^q = h^q \in \langle h angle$. This shows that $H \subseteq \langle h angle$. Since $h \in H$, $\langle h angle \subseteq  H$, therefore 
$$H = \langle h angle = \langle g^r angle$$
is cyclic of order $k$.

\bigskip
It remains to prove that $H = H_0$. Since $g^r$ has order $k$, $g^{rk} = e$, where $g$ is a generator of $G$, so $n \mid rk$. Then $q = \frac{n}{k} \mid r$, and $r = uq$ for some integer $u$.  Therefore $g^r = (g^q)^u \in \langle g^q angle = H_0$, thus $H = \langle g^r angle \subseteq H_0$,  where $|H_0| = |H| = k$. We can conclude that $H = H_0$.

There is exactly one subgroup of order $k$ for each divisor $k$ of $n$.

\end{proof}

Exercise 10.3.5 (Unique subgroup of cyclic group)

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Exercise 10.3.5 (Unique subgroup of cyclic group)

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